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The opamp here is ideal except for the bias currents. In order to calculate the maximum absolute voltage deviation at the output, I was thinking of sum the deviation of Vos and deviation of current, being that the deviation of current, for the maximum is R2*Ib (I think.)

My problem is for the deviation of Vos because in the exercise description, they give Ios (so I did the Ohm laws to get Vos) and then calculate the deviation for Vos with gain*Vos, but then I got the wrong answer in the end. What am I doing wrong?

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    \$\begingroup\$ Show your calculations please. \$\endgroup\$
    – Andy aka
    Commented May 29, 2023 at 18:53
  • \$\begingroup\$ !Valid XHTML \$\endgroup\$
    – Patricia Vieira
    Commented May 29, 2023 at 19:18
  • \$\begingroup\$ Ib(+)=7.5e-9 + (1/2)(1.4e-9)=8.2e-9 gain=R2(cR1=248.28e6 -Current deviation: Vo(Ib)=R2*Ib(+)=5.33e-5 -Voltage deviation: Vo(Vos)=gain*(IosR2)=2.25e12 -Maximum absolute voltage deviation=2.25e12 V( which is very wrong) \$\endgroup\$
    – Patricia Vieira
    Commented May 29, 2023 at 19:22
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    \$\begingroup\$ You need to put it in your question and, if possible, format it using latex. Eg \$I_{b+} = 7.5\text{ nA}+\dfrac{1.4\text{ nA}}{2}\$ is written like this: \$I_{b+} = 7.5\text{ nA}+\dfrac{1.4\text{ nA}}{2}\$ \$\endgroup\$
    – Andy aka
    Commented May 29, 2023 at 19:38

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